ACSI 2022 Y6 HL Mathematics Prelim Exam Paper 2 (Student Solutions)

ACS (Independent) / Mathematics Department / Mathematics AA HL / Year 6 / 2022 Preliminary Examinations / Paper 2 / Solutions 1 PRELIM 2022 PAPER 2 YEAR 6 IB DIPLOMA PROGRAMME MATHEMATICS : ANALYSIS AND APPROACHES HIGHER LEVEL STUDENT SOLUTIONS SECTION A Qn Solution COMMENTS 1. [4 marks] (a) Area of the minor segment = Area of the sector – Area of OAB 2 21 1 sin2 2r r   Some mistook sector for segment. (b)   2 2 2 2 2 1 1 1 sin2 2 4 2 sin 2 2sin (shown) r r r r r                  This is correct if (a) is correct O A B

ACS (Independent) / Mathematics Department / Mathematics AA HL / Year 6 / 2022 Preliminary Examinations / Paper 2 / Solutions 2 Qn Solution COMMENTS (c) 2.30988 2.31 (3sf)  Many could not do. Must use Graph or nsolve 2. [4 marks] ' sin cosf x x x Observe that  cos sind x xdx  .     1 2 3 2 3 2 sin cos ( sin ) cos cos 3 2 2 cos3 f x x x dx x x dx x c x c          Substituting , 0 :3x y  3 2 3 2 20 cos3 3 2 1 1 3 2 3 2 c c            Thus in exact form we have    3 22 1cos3 3 2f x x  Many did not see this as in the form of ∫𝑓ᇱ(𝑥)[𝑓(𝑥)]௡𝑑𝑥=[௙(௫)]೙శభ ௡ାଵ +𝑐 Not acceptable for decimal value of ଵ ଷ√ଶ. All other surd forms of ଵ ଷ√ଶ are accepted. 3. [6 marks]

ACS (Independent) / Mathematics Department / Mathematics AA HL / Year 6 / 2022 Preliminary Examinations / Paper 2 / Solutions 3 Qn Solution COMMENTS (a)    0.95 0.6 0.7 0.35 P X Y P X P Y P X Y P X Y P X Y              Most got it correct (b)    0.35 0.6 7 12 P X YP Y X P X    ଻ ଵଶ is an exact value so no need to convert to 0.583. Some students only gave as 0.58. (c) 𝑃(𝑋∪𝑌ᇱ)=0.25+0.35+0.05=0.65 Alternatively, accept ' 1 1 0.7 0.35 0.65 P X Y P Y P X Y         Badly done. 4. [7 marks]

ACS (Independent) / Mathematics Department / Mathematics AA HL / Year 6 / 2022 Preliminary Examinations / Paper 2 / Solutions 4 Qn Solution COMMENTS (a) From GDC we have Mean 50.3 Variance   22 23.7152 562.4107 562 (3sf)    Key in data and use one-var stats. Scroll down and the standard deviation is 23.7152 so need to square it to obtain the variance. Please correct to 3 significant figures (b) From GDC we have 33, 51, 73a b c   (c) New mean 50.3new q   New standard deviation 23.7 (3sf)new  Standard deviation measures the spread of the data so it is still the same if every datum is decreased by the same value. 5. [7 marks] (a) ln 1 v A t B dv dt t B       Substituting 110, : 20 dvt dt  1 1 20 10B   10B  Substituting 100, 0 :t v    0 ln 100 10 ln 1